Question

Differentiate
${e}^{2x}$
from first principles​

Answer 1

AnswEr :

Given$\sf \: f(x) = {e}^{2x}$

Using first principle,

$\implies \sf \: f'(x) = \lim_ {h \longrightarrow \: 0} \sf \dfrac{ {e}^{2(x \: + h)} - {e}^{2x} }{h} \\ \\ \implies \sf f'(x) =\lim_ {h \: \longrightarrow 0} \: \dfrac{ {e}^{2x}. {e}^{2h} - {e}^{2x} }{h} \\ \\ \implies \sf \: f'(x) = \lim_ {h \: \longrightarrow 0}\: \dfrac{ {e}^{2x}({e}^{2h} - 1) }{h} \\ \\ \implies \sf \: f'(x) ={e}^{2x} \lim_ {h \: \longrightarrow 0} \: \dfrac{({e}^{2h} - 1) }{h}$

Multiplying and Dividing by 2,

$\implies \sf \: f'(x) = 2{e}^{2x}\lim_ {2h \: \longrightarrow 0} \: \dfrac{({e}^{2h} - 1) }{2h}$

Using the formula,

$\sf \lim_ {x \longrightarrow \: 0} \: \dfrac{ {e}^{x} - 1}{x} = 1$

Thus,

$\implies \sf \: f'(x) = 2{e}^{2x} \: \times 1 \\ \\ \implies \boxed{ \boxed{ \sf f'(x) = 2{e}^{2x}}}$